Answer
**Part (a):**
**i.**
Given the equations:
\[ x^{2} + y^{A^{2}} = 4 \]
\[ y = 2 \]
Substituting \( y = 2 \) into the first equation:
\[ x^{2} + (2)^{A^{2}} = 4 \]
\[ x^{2} = 4 - 2^{A^{2}} \]
\[ x = \pm \sqrt{4 - 2^{A^{2}}} \]
**ii.**
Given the equations:
\[ x^{2} + y^{7} = 4 \]
\[ y = x + 1 \]
Substituting \( y = x + 1 \) into the first equation:
\[ x^{2} + (x + 1)^{7} = 4 \]
\[ x^{2} + x^{7} + 7x^{6} + 21x^{5} + 35x^{4} + 35x^{3} + 21x^{2} + 7x + 1 = 4 \]
\[ x^{7} + 7x^{6} + 21x^{5} + 35x^{4} + 35x^{3} + 22x^{2} + 7x - 3 = 0 \]
This is a 7th-degree polynomial equation, which is complex to solve analytically. Numerical methods or graphing techniques would be required to approximate the solutions.
**Part (b):**
The solutions to the system of equations in part (a) represent the points of intersection between the two curves defined by the equations. Specifically:
- In part (a)i, the solutions \( x = \sqrt{4 - 2^{A^{2}}} \) and \( x = -\sqrt{4 - 2^{A^{2}}} \) indicate the points where the hyperbola-like curve \( x^{2} + y^{A^{2}} = 4 \) intersects the horizontal line \( y = 2 \).
- In part (a)ii, the solutions to the 7th-degree polynomial equation represent the points where the 7th-degree curve \( x^{2} + y^{7} = 4 \) intersects the straight line \( y = x + 1 \).
These intersection points are significant in various applications, such as determining equilibrium points in physics, solving optimization problems in economics, or analyzing geometric intersections in engineering.
Solution
Alright, I need to solve the given system of equations and then interpret the results. Let's tackle part (a) first.
**Part (a):**
**i. \( x^{2} + y^{A^{2}} = 4 \) and \( y = 2 \)**
First, I'll substitute \( y = 2 \) into the first equation to find the value of \( x \).
So, the equation becomes:
\[ x^{2} + (2)^{A^{2}} = 4 \]
Now, I need to solve for \( x \). However, there's a variable \( A \) in the exponent of \( y \), which complicates things. Since \( A \) is not defined, I'll assume it's a constant for the sake of solving. Let's denote \( A^{2} = k \), where \( k \) is a constant.
So, the equation simplifies to:
\[ x^{2} + 2^{k} = 4 \]
Subtracting \( 2^{k} \) from both sides:
\[ x^{2} = 4 - 2^{k} \]
Taking the square root of both sides:
\[ x = \sqrt{4 - 2^{k}} \]
But since \( x \) can be positive or negative, the solutions are:
\[ x = \sqrt{4 - 2^{k}} \quad \text{and} \quad x = -\sqrt{4 - 2^{k}} \]
However, without a specific value for \( k \) (which depends on \( A \)), I can't provide numerical solutions. If \( A \) is given, I can substitute it back to find the exact values of \( x \).
**ii. \( x^{2} + y^{7} = 4 \) and \( y = x + 1 \)**
Again, I'll substitute \( y = x + 1 \) into the first equation to eliminate one variable.
Substituting:
\[ x^{2} + (x + 1)^{7} = 4 \]
Now, I need to solve for \( x \). Expanding \( (x + 1)^{7} \) using the binomial theorem:
\[ (x + 1)^{7} = x^{7} + 7x^{6} + 21x^{5} + 35x^{4} + 35x^{3} + 21x^{2} + 7x + 1 \]
So, the equation becomes:
\[ x^{2} + x^{7} + 7x^{6} + 21x^{5} + 35x^{4} + 35x^{3} + 21x^{2} + 7x + 1 = 4 \]
Bringing all terms to one side:
\[ x^{7} + 7x^{6} + 21x^{5} + 35x^{4} + 35x^{3} + 22x^{2} + 7x - 3 = 0 \]
This is a 7th-degree polynomial equation, which is quite complex to solve analytically. Typically, such equations are solved numerically using methods like Newton-Raphson or graphing techniques to approximate the roots. Without specific tools or further information, I can't provide exact solutions here.
**Part (b): What does your answer in (a) represent?**
The solutions to the system of equations in part (a) represent the points of intersection between the two curves defined by the equations. Specifically:
- In part (a)i, the equation \( x^{2} + y^{A^{2}} = 4 \) represents a hyperbola-like curve, and \( y = 2 \) is a horizontal line. The solutions \( x = \sqrt{4 - 2^{k}} \) and \( x = -\sqrt{4 - 2^{k}} \) indicate the points where the hyperbola intersects the horizontal line \( y = 2 \).
- In part (a)ii, the equation \( x^{2} + y^{7} = 4 \) represents a curve that is more complex due to the exponent 7, and \( y = x + 1 \) is a straight line with a slope of 1 and a y-intercept of 1. The solutions to this system would represent the points where the 7th-degree curve intersects the straight line \( y = x + 1 \).
These intersection points are significant in various applications, such as finding equilibrium points in physics, determining optimal solutions in economics, or solving geometric problems where two shapes intersect.
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